# Boundary conditions for eigenfunctions in a potential step

1. Homework Statement
A particle with mass m and spin 1/2, it is subject in a spherical potencial step with height ##V_0##.
How is the general form for the eigenfunctions?
What is the boundary conditions for this eigenfunctions?
Find the degeneracy level for the energy, when it is ##E<V_0##

2. Homework Equations

-\frac{\hbar^2}{2m}\left(\frac{\partial^2u}{\partial r^2}-\frac{l(l+1)}{r^2}u\right)+V(r)u(r)=Eu(r)

with ##u(r)=rR(r)##

3. The Attempt at a Solution
Well, first the constant of motion it ##\hat{L_z}##, ##\hat{L^2}## and ##\hat{S_z}## because conmute with ##\hat{H}##, also ##\hat{H}##, and it former a complete set of observables that conmute, therefore the form of eigenfunctions are:

\psi(r,\theta,\varphi,\sigma)=R(r)Y(\theta,\varphi)\chi(\sigma)

when ##\chi(\sigma)## is a spin function, and only take the values ##m_s={+1/2,-1/2}## because eigenvalues of operator ##\hat{S^2}## is ##3/4\hbar^2## with ##s=1/2##.

Here my doubt:
Boundary conditions to ##Y(\theta,\varphi)## this is referred to:
##l=0, 1, \dots, n-1## and ##-l\leq m\leq +l##?

For radial function, it must be:
because ##\hat{H}## is hermitian, then for ##r=0\Longrightarrow u(0)\longrightarrow0##
and for ##r\longrightarrow\infty \Longrightarrow u\longrightarrow0##

For ##\chi## the boundary conditions it is ##-s\leq m_s\leq+s##?

In the case for the degeneracy, when ##E<0## corresponds to ##l=0## because it is a bound state, then without spin the level of degeneracy is ##n^2## but with spin is ##2n^2##.

Up to here my poor attemp to solution, i am really puzzled for explain the boundary condition to Y and ##\chi##, and i am not sure that the
##r\longrightarrow\infty \Longrightarrow u\longrightarrow0## for this potential.

Thanks a lot!